Question

find the projection of u along v
u=<-1,2,0> and v=<2,0,1>

Answer 1

the formula for projection:

\[\frac{ u \cdot v }{ \left| v \right|^2 }(v)\]

Answer 2

so
\[{ u \cdot v }\] =
(-1(2))+2(0)+(0)1) = -2

the magnitude of |v|
\[\sqrt{2^2 + 0^2+1^2}\]=\[\sqrt{5}\]

so
\[\frac{ -2 }{ \sqrt{5} }<2,0,1>\]

Answer 3

http://www.youtube.com/watch?v=um5QM1uRPHM

http://www.youtube.com/watch?v=FvcHKhdaAyw